Math 241 Final Review Sheet
Information about the final: The final exam is scheduled for ... The final is worth two hour
exams. It will consist of 12 questions. 6 questions will cover sections 12.9 through 13.7. The other
six questions will be about the rest of the course. You will be allowed to bring an 8.5 × 11
sheet of paper with whatever you wish written on each side..
About this review sheet: This review sheet covers sections 12.9-13.7 only. Please go over the
old review sheets, exams and homeworks to review for the other part of the test.
1. Evaluate the integral
!!
R
(x2 − xy + y 2 ) dA,
where R is the region bounded by the ellipse
x2 − xy + y 2 = 2,
"
"
√
√
by using the transformation x = 2u − 2/3v; y = 2u + 2/3v.
Solution: The
step is to figure √
out what
Plug" happens to R under this2 transformation.
√ first "
2
ging in x = 2u − 2/3v and y = 2u + 2/3v into the equation x − xy + y = 2, after a
bit of algebra the equation becomes u2 + v 2 = 1. That’s why the transformation the problem
suggests is so useful. The ellipse R has mapped onto the unit circle. Next calculate the
Jacobian
#
#
∂(x, y) ## xu xv ##
=
∂(u, v) # yu yv #
"
# √
#
# 2 − 2/3 #
#
#
"
√
=#
2
2/3 #
√ "
= 2 2 2/3
√
= 4/ 3
√
4 3
=
3
%
&
√
$$
So our integral becomes D (u2 + v 2 ) 4 3 3 dA where D is the region u2 + v 2 ≤ 1. A good
way to evaluate this integral is polar coordinates. The unit disk, D, is parameterized by
0 ≤ θ2π and 0 ≤ r ≤ 1. The function u2 + v 2 becomes r2 . So we get
! 2π ! 1 ' √ (
4 3
r2
r dr dθ.
3
0
0
This evaluates to
√
2 3π
.
3
2. Match the vector fields with their plots. Each plot is centered at (0, 0) .
1
(1) %y, 0&
(A)
(B)
(C)
(D)
(E)
(F )
(2) %0, x&
(3) %−xy, x&
(4) %x2 − y 2 , 1&
(5) %sin(4x), 0&
(6) %−y, x&
Solution: There is only one picture where the x component of all the vectors is 0, namely (B).
Hence (B) matches (2). Two of the pictures have the y component 0, namely (D) and (E).
So we have to figure out which one matches (1) and which one matches (5). The equation for
(1) vanishes when y = 0, so the vector field is zero along the x-axis. Hence (1) matches (E),
and so (5) must match (D).
Continuing, one of the most notable features of (F) is that it is zero along the y axis. Of the
remaining equations (3),(4) and (6) only one vanishes along the y-axis, namely (3). (Plug in
x = 0 and you get %0, 0&.)
Equation (4) always has y component 1. This matches picture (C), but definitely not (A).
So the only one left to match (A) is (6).
Here is a summary of the answers: A6, B2, C4, D5, E1, F3.
3. Evaluate the line integral
!
C
z 2 dx − z dy + 2y dz,
where C consists of the line segments from (0, 0, 0) to (0, 1, 1) and (0, 1, 1) to (1, 2, 3).
Solution: We should parameterize the two line segments first of all. For example
$r1 (t) = %0, t, t&, 0 ≤ t ≤ 1
$r2 (t) = %t, 1 + t, 1 + 2t&, 0 ≤ t ≤ 1
2
are perfectly good parameterizations.
Let’s concentrate on the first line segment. Differentiating the components of $r1 with respect
to t, dx = 0 dt, dy = dt, dz = dt. So we get
! 1
! 1
1
2
t (0) dt − t dt + 2t dt =
t dt = .
2
0
0
Now we attack the second line segment. Differentiating the components of $r2 with respect to
t, we get dx = dt, dy = dt, dz = 2 dt. So we get
! 1
! 1
16
2
(1 + 2t) dt − (1 + 2t) dt + 2(1 + t)2 dt =
4t2 + 4t + 2 dt = .
3
0
0
Adding up the results for each line segment, we get
1
2
+
16
3
=
35
6 .
4. A 160 − lb man carries a 25 − lb can of paint up a helical staircase that encircles a silo with
a radius of 20 − f t. If the silo is 90 − f t high and the man makes exactly three complete
revolutions, how much work is done by the man against gravity in climbing to the top?
$
Solution: Work along a path C is defined to be C F$ · d$r. The force the man has to exert
against gravity at each point is F$ = %0, 0, 185&.
One way to do the problem is to parameterize C and actually do the integral. However, the
field F$ is conservative. It has potential function f (x, y, z) = 180z. (Verify that ∇f = F$ .) So
we can use the fundamental theorem of line integrals
!
F$ · d$r = f (end point) − f (starting point).
C
We can take the starting point to be (0, 0, 0) and the finish to be (0, 0, 90). (The silo is 90
feet high.) So we get that the work is (180)(90) − (180)(0) = 16, 200 ft-lbs.
5. Show that the line integral
!
C
2x sin y dx + (x2 cos y − 3y 2 ) dy
is independent of path. Evaluate the integral along a path from (−1, 0) to (5, 1).
Solution: To see that it is independent of path, we must check that Qx = Py . But this is
true. Both equal 2x cos y. Now suppose we have ∇f = %2x sin y, x2 cos y − 3y 2 &. Then
fx = 2x sin y
!
f = 2x sin y dx + g(y)
f = x2 sin y + g(y)
Now differentiate f with respect to y, and set it equal to Q = x2 cos y − 3y 2
fy = x2 cos y + g " (y)
= x2 cos y − 3y 2
So g " (y) = −3y 2 , and g(y) = −y 3 . Thus
f (x, y) = x2 sin y − y 3 .
Then the line integral from (−1, 0) to (5, 1) is just f (5, 1) − f (−1, 0) = 25 sin(1) − 1.
3
6. Use Green’s theorem to evaluate the line integral
!
y 3 dx − x3 dy
C
where C is the circle x2 + y 2 = 4.
Solution: Green’s theorem says that the above line integral is equal to
!!
Qx − Py dA,
D
where D is the disk of radius 2 centered at (0, 0). Now Qx − Py = −3x2 − 3y 2 = −3(x2 + y 2 ).
It makes sense to use polar coordinated to evaluate this integral. Qx − Py becomes −3r2 . So
we get
!
!
2π
0
2
0
(−3r2 )r dr dθ = −24π
7. Let f be a function and F$ a vector field. State whether each expression is meaningful. If so,
state whether it is a function or a vector field.
(a) curl f Not meaningful.
(b) div F$ Function.
(c) grad f Vector field.
(d) curl(grad f ) Vector field.
(e) grad F$ Not meaningful.
(f) grad(div F$ ) Vector field.
(g) div(grad f ) Function.
(h) grad(div f ) Not meaningful
(i) curl(curl F$ ) Vector field.
(j) div(div F$ ) Not meaningful.
(k) (grad f ) × (div F$ )Not meaningful.
(l) div(curl(grad f )) Function.
8. Find the curl and divergence of the vector field
F$ (x, y, z) = %xyz, −x2 y, 2x&.
Solution: div F$ = yz − x2 and curl F$ = %0, −(2 − xy), −2xy − xz&
9. Evaluate the surface integral
!!
yz dS,
S
where S is the surface with parametric equations x = uv, y = u + v, z = u − v, u2 + v 2 ≤ 1.
4
Solution: First we calculate the correction term |$ru ×$rv |. So $ru = %v, 1, 1& and $rv = %u, 1, −1&.
Now
#
#
# $i $j $k #
#
#
|$ru × $rv | = ## v 1 1 ##
# u 1 −1 #
= |%−2, −(−v − u), v − u&|
"
= 4 + (u + v)2 + (v − u)2
"
= 4 + 2(u2 + v 2 )
So our integral is
!!
!!
"
"
(u + v)(u − v) 4 + 2(u2 + v 2 ) dA =
(u2 − v 2 ) 4 + 2(u2 + v 2 ) dA
D
D
where D is the disk
+
≤ 1. Let’s evaluate this using polar coordinates: u2 + v 2 = r2 ,
u = r cos θ and v = r sin θ.
! 2π ! 1
! 2π ! 1
%"
&
"
2
2
2
2
2
(r cos θ − r sin θ)
4 + 2r r dr dθ =
(cos2 θ − sin2 θ)r3 4 + 2r2 dr dθ
0
0
0
0
! 2π
! 1 "
=
cos2 θ − sin2 θ dθ
r3 4 + 2r2 dr
u2
v2
0
0
$ 2π
Let’s first calculate 0 cos2 θ − sin2 θ dθ. This is zero. One way to see this is to argue that
sin2 θ is the same as cos2 θ only
right. Hence the area under one complete cycle
$ 2π shifted to the
$ 2π
is going to be the same. So 0 sin2 θ dθ = 0 cos2 θ dθ, and the total integral cancels to be
zero. But then
!!
! 1 "
yz dS = (0)
r3 4 + 2r2 dr = 0,
S
0
and there’s no need to evaluate the second integral.
$$
$ where
10. Use Stokes’ theorem to evaluate S curl F$ · dS,
F$ (x, y, z) = %x2 eyz , y 2 exz , z 2 exy &
and S is the hemisphere x2 + y 2 + z 2 = 4 z ≥ 0, oriented upward.
Solution: According to Stokes’ theorem, this integral can be calculated as
!
F$ · d$r,
C
where C is the boundary of S, which is the circle in the xy-plane x2 + y 2 = 4. Now
F$ · d$r = %x2 eyz , y 2 exz , z 2 exy & · %dx, dy, dz&
= x2 eyz dx + y 2 exz dy + z 2 exy dz
Now we need to parameterize C. A good way is using the angle θ. So x = 2 cos θ, y = 2 sin θ
and z = 0, 0 ≤ θ ≤ 2π. So
dx = −2 sin θ dθ
dy = 2 cos θ dθ
dz = 0 dθ
5
So our integral becomes
! 2π
(2 cos θ)2 e(2 sin θ)0 (−2 sin θ) dθ + (2 sin θ)2 e(2 cos θ)0 (2 cos θ) dθ + 02 e(2 cos θ)(2 sin θ) 0 dθ
0
! 2π
=8
sin2 θ cos θ − sin θ cos2 θ dθ
0
)! 2π
*
! 2π
2
2
=8
sin θ cos θ dθ −
cos θ sin θ dθ
0
0
We can solve
$ 0 the first integral by substituting$u2π= sin θ. Then du = cos θ dθ, and the integral
becomes 0 u2 du = 0. Similarly the integral 0 cos2 θ sin θ is zero. So the final answer is 0.
$$
$ that is calculate the
11. Use the divergence theorem to calculate the surface integral S F$ · dS;
$
flux of F across S.
F$ (x, y, z) = %ex sin y, ex cos y, yz 2 & and S is the surface of the box bounded by the planes
x = 0, x = 1, y = 0, y = 1, z = 0, z = 2.
Solution: According to the divergence theorem
!!
!!!
$=
F$ · dS
div F$ dV
S
E
where E is the solid region bounded by the box S. So
E = {(x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 2.
Also
div F$ = Px + Qy + Rz
= ex sin y − ex sin y + 2yz
= 2yz
So our integral is
!
0
2! 1! 2
0
2yz dx dy dz = 2
0
)!
0
=2
6
1
* )!
dx
0
1
y dy
* )!
0
2
z dz
*